Re: Super densely ordered sets

On Sun, 17 Feb 2008, Denis Feldmann wrote:

On Feb 15, 12:27 pm, marc...@xxxxxxxxx wrote:

Let S be infinite set and > be a strict linear order on S. Let say
that S is super densely ordered by > iff:

For all subsets X and Y of S with |X| <= Aleph_0 and |Y| <= Aleph_0
such that a < b for all a in X, all b in Y, there exists c in S such
that a < c < b for all a in X, all b in Y.

Do super densely ordered sets exist?

If S is cofinal to aleph_0, then S isn't super dense.
Let X be a cofinal set and Y empty.

A much simpler construction (equivalent ?) was discussed here a few
years ago : the idea is to use analog of the binary construction of
reals (as sequences of 0 and 1) : we take the lexicographic order on the
sequences of {0,1]^alpha (where alpha is a cardinal aleph _x), the trick
being of using only those sequences such that there is a *last* 1, i.e
an ordinal beta <alpha such that s_beta =1 and for all beta'>beta,
s_beta'=0.

Because of the above, aleph_xi cannot be cofinal to aleph_0.
In particular, xi /= 0, omega_0.

To detail your 'last one', you are requiring that there is at
least one 1. Otherwise the zero sequence O, is bottom element and

S = lex { 0,1 }^aleph_xi restricted to 0 terminating sequences

isn't superdense for letting X be empty and Y = { O }.

.

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